I've been working on some old prelims from my university when they used to just be on point-set topology. We don't cover a couple of the topics so I've been teaching myself some of the material, one of the topics no longer covered is Stone-Čech Compactification. Which I have a somewhat tenuous understanding of, at any rate one of the questions reads:

Let $X$ be a completely regular topological space and let $\beta(X)$ denote the Stone-Čech compactification of $X$. Show that every $y \in \beta(X) \setminus X$ is a limit point of $X,$ but is not the limit of a sequence of points in $X$.

It's clear to me how to go about the first part, $X$ if considered as a subset of $\beta(X)$ is dense in $\beta(X)$. Then it follows that for every point $y \in \beta(X) \setminus X$ that every neighborhood of $y, U$ in $\beta(X)$ will touch $X$.

For the second part of the question, I must say, sadly, that I'm at a loss in general. Presumably we need to assume that we have some convergent sequence $\{x_i\}_{i \in \mathbb{N}}$ that converges to a point $y \in \beta(X) \setminus X$ and show that this is a contradiction. But, probably due to my weak understanding of the Stone-Čech Compactification I am unsure of how to go about this.

  • $\begingroup$ How did you define the S-C compactification? $\endgroup$ – Asaf Karagila Apr 27 '11 at 4:49
  • $\begingroup$ I'm going off Munkres' definition who defines it via closure of an embedding of $X$ into some product of $[0,1]$. Essentially taking all continuous $f_\lambda: X \rightarrow [0,1]$ and defining a function $h: X \rightarrow \prod_{\lambda \in \Lambda} [0,1]_\lambda$ by $h(x)=(f_\lambda(x))_{\lambda \in \Lambda}$ so that $\beta(x)$ is the closure of $h(X)$. $\endgroup$ – JSchlather Apr 27 '11 at 5:06
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    $\begingroup$ I'm not sure how to answer this exactly (nor I have the time at the moment), the smidgen of intuition I can give you is that $\beta(X)$ is much too big for sequences, for example $|\mathbb N|=\aleph_0$ and $|\beta(\mathbb N)|=2^{2^{\aleph_0}}$. $\endgroup$ – Asaf Karagila Apr 27 '11 at 8:11
  • $\begingroup$ For tha case that X is discrete countable space, it is Corollary 3.6.15 in Engelking's Genral topology. (In case you have the access to this book it might help.) See also here: thales.doa.fmph.uniba.sk/sleziak/texty/rozne/engel/engel.pdf The same result is shown here books.google.com/… $\endgroup$ – Martin Sleziak Apr 27 '11 at 10:57
  • $\begingroup$ The result about $\beta\omega$ is also shown in this answer on MO. $\endgroup$ – Martin Sleziak Jul 30 '13 at 12:06

EDIT: As pointed by Stefan H. in his comment, the solution I have suggested only works if $X$ is normal, since I am using Tietze extension theorem.

Perhaps I have overlooked something and I will be blushing, but I will give it a try. (This is my solution, I did not check the books I mentioned in the above comment. Perhaps the proofs from those books can give you a hint for a different proof.)

Let $x_n\in X$ be a sequence which converges to $x\in\beta X\setminus X$. We can assume that $x_n$'s are distinct. I will show bellow that $\{x_n; n\in\mathbb N\}$ is closed discrete subspace of $X$. But first I will show how to use this fact.

For any choice of $y_n\in[0,1]$, $n\in\mathbb N$, we can define $f(x_n)=y_n$ and extend it continuously (by Tietze's theorem) to the whole $X$. Now there exists a continuous extension $\overline f : \beta X \to [0,1]$. By continuity, the sequence $y_n=\overline f(x_n)$ converges to $\overline f(x)$. We have shown that every sequence in $[0,1]$ is convergent, a contradiction.

Now to the proof that $\{x_n; n\in\mathbb N\}$ is closed and discrete.

Since $\{x_n; n\in\mathbb N\}\cup\{x\}$ is a compact subset of $\beta X$, it is closed in $\beta X$. The intersection with $X$ is $\{x_n; n\in\mathbb N\}$ and it must be closed in $X$.

Now we consider $\{x_n; n\in\mathbb N\}$ as a subspace of $X$ and we want show that this subspace is discrete. Choose some $x_n$. By Hausdorffness, it can be separated from $x$, i.e. there exists a neighborhood $U\ni x$ such that $x_n\notin U$ and a neighborhood $V\ni x_n$ with $V\cap U=\emptyset$.

Now by convergence $U$ contains all but finitely many $x_n$'s, hence using Hausdorfness we can separate $x_n$ from the (finitely many) remaining ones.

  • $\begingroup$ I have a question: How can you extend the function from the sequence to the whole $X$ if $X$ is not normal? $\endgroup$ – Stefan Hamcke Sep 21 '13 at 10:44
  • $\begingroup$ @StefanH. Good point. I've edit my answer to point out that my answer works only for normal space (at least in the way it is formulated at this moment). I hope I will get back to this problem and think a little more about it when I have more time. $\endgroup$ – Martin Sleziak Sep 21 '13 at 12:32
  • $\begingroup$ My guess is that you did it almost exactly the way students were supposed to. For details see my answer below. $\endgroup$ – Niels J. Diepeveen Aug 23 '15 at 14:58

In Gillman & Jerison's Rings of continuous functions I found the following note

8.21 N.B. A number of authors have fallen into the trap of assuming then every countable, closed, discrete subset of a completely regular space is $C^\ast$-embedded. We have just seen a counterexample: [...]

It seems likely that one of these authors, or someone following them has made up this problem, as the same counterexample works here. (Martin Sleziak's answer shows how the assumption would lead to a proof).

The counterexample referred to is the Tychonoff plank $([0, \omega_1] \times [0, \omega]) \setminus (\omega_1, \omega)$. It can be shown that every real-valued function on that space can be extended to its one-point compactification $[0, \omega_1] \times [0, \omega]$. That implies that this is also the Stone-Čech compactification, and clearly $\lim_{n\to\infty} (\omega_1, n) = (\omega_1, \omega)$.


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